2.222 222 222 222 223 88 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.222 222 222 222 223 88(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
2.222 222 222 222 223 88(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 2.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.222 222 222 222 223 88.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.222 222 222 222 223 88 × 2 = 0 + 0.444 444 444 444 447 76;
  • 2) 0.444 444 444 444 447 76 × 2 = 0 + 0.888 888 888 888 895 52;
  • 3) 0.888 888 888 888 895 52 × 2 = 1 + 0.777 777 777 777 791 04;
  • 4) 0.777 777 777 777 791 04 × 2 = 1 + 0.555 555 555 555 582 08;
  • 5) 0.555 555 555 555 582 08 × 2 = 1 + 0.111 111 111 111 164 16;
  • 6) 0.111 111 111 111 164 16 × 2 = 0 + 0.222 222 222 222 328 32;
  • 7) 0.222 222 222 222 328 32 × 2 = 0 + 0.444 444 444 444 656 64;
  • 8) 0.444 444 444 444 656 64 × 2 = 0 + 0.888 888 888 889 313 28;
  • 9) 0.888 888 888 889 313 28 × 2 = 1 + 0.777 777 777 778 626 56;
  • 10) 0.777 777 777 778 626 56 × 2 = 1 + 0.555 555 555 557 253 12;
  • 11) 0.555 555 555 557 253 12 × 2 = 1 + 0.111 111 111 114 506 24;
  • 12) 0.111 111 111 114 506 24 × 2 = 0 + 0.222 222 222 229 012 48;
  • 13) 0.222 222 222 229 012 48 × 2 = 0 + 0.444 444 444 458 024 96;
  • 14) 0.444 444 444 458 024 96 × 2 = 0 + 0.888 888 888 916 049 92;
  • 15) 0.888 888 888 916 049 92 × 2 = 1 + 0.777 777 777 832 099 84;
  • 16) 0.777 777 777 832 099 84 × 2 = 1 + 0.555 555 555 664 199 68;
  • 17) 0.555 555 555 664 199 68 × 2 = 1 + 0.111 111 111 328 399 36;
  • 18) 0.111 111 111 328 399 36 × 2 = 0 + 0.222 222 222 656 798 72;
  • 19) 0.222 222 222 656 798 72 × 2 = 0 + 0.444 444 445 313 597 44;
  • 20) 0.444 444 445 313 597 44 × 2 = 0 + 0.888 888 890 627 194 88;
  • 21) 0.888 888 890 627 194 88 × 2 = 1 + 0.777 777 781 254 389 76;
  • 22) 0.777 777 781 254 389 76 × 2 = 1 + 0.555 555 562 508 779 52;
  • 23) 0.555 555 562 508 779 52 × 2 = 1 + 0.111 111 125 017 559 04;
  • 24) 0.111 111 125 017 559 04 × 2 = 0 + 0.222 222 250 035 118 08;
  • 25) 0.222 222 250 035 118 08 × 2 = 0 + 0.444 444 500 070 236 16;
  • 26) 0.444 444 500 070 236 16 × 2 = 0 + 0.888 889 000 140 472 32;
  • 27) 0.888 889 000 140 472 32 × 2 = 1 + 0.777 778 000 280 944 64;
  • 28) 0.777 778 000 280 944 64 × 2 = 1 + 0.555 556 000 561 889 28;
  • 29) 0.555 556 000 561 889 28 × 2 = 1 + 0.111 112 001 123 778 56;
  • 30) 0.111 112 001 123 778 56 × 2 = 0 + 0.222 224 002 247 557 12;
  • 31) 0.222 224 002 247 557 12 × 2 = 0 + 0.444 448 004 495 114 24;
  • 32) 0.444 448 004 495 114 24 × 2 = 0 + 0.888 896 008 990 228 48;
  • 33) 0.888 896 008 990 228 48 × 2 = 1 + 0.777 792 017 980 456 96;
  • 34) 0.777 792 017 980 456 96 × 2 = 1 + 0.555 584 035 960 913 92;
  • 35) 0.555 584 035 960 913 92 × 2 = 1 + 0.111 168 071 921 827 84;
  • 36) 0.111 168 071 921 827 84 × 2 = 0 + 0.222 336 143 843 655 68;
  • 37) 0.222 336 143 843 655 68 × 2 = 0 + 0.444 672 287 687 311 36;
  • 38) 0.444 672 287 687 311 36 × 2 = 0 + 0.889 344 575 374 622 72;
  • 39) 0.889 344 575 374 622 72 × 2 = 1 + 0.778 689 150 749 245 44;
  • 40) 0.778 689 150 749 245 44 × 2 = 1 + 0.557 378 301 498 490 88;
  • 41) 0.557 378 301 498 490 88 × 2 = 1 + 0.114 756 602 996 981 76;
  • 42) 0.114 756 602 996 981 76 × 2 = 0 + 0.229 513 205 993 963 52;
  • 43) 0.229 513 205 993 963 52 × 2 = 0 + 0.459 026 411 987 927 04;
  • 44) 0.459 026 411 987 927 04 × 2 = 0 + 0.918 052 823 975 854 08;
  • 45) 0.918 052 823 975 854 08 × 2 = 1 + 0.836 105 647 951 708 16;
  • 46) 0.836 105 647 951 708 16 × 2 = 1 + 0.672 211 295 903 416 32;
  • 47) 0.672 211 295 903 416 32 × 2 = 1 + 0.344 422 591 806 832 64;
  • 48) 0.344 422 591 806 832 64 × 2 = 0 + 0.688 845 183 613 665 28;
  • 49) 0.688 845 183 613 665 28 × 2 = 1 + 0.377 690 367 227 330 56;
  • 50) 0.377 690 367 227 330 56 × 2 = 0 + 0.755 380 734 454 661 12;
  • 51) 0.755 380 734 454 661 12 × 2 = 1 + 0.510 761 468 909 322 24;
  • 52) 0.510 761 468 909 322 24 × 2 = 1 + 0.021 522 937 818 644 48;
  • 53) 0.021 522 937 818 644 48 × 2 = 0 + 0.043 045 875 637 288 96;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.222 222 222 222 223 88(10) =

0.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 1011 0(2)

5. Positive number before normalization:

2.222 222 222 222 223 88(10) =

10.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 1011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.222 222 222 222 223 88(10) =

10.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 1011 0(2) =

10.0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 1011 0(2) × 20 =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 10(2) × 21


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 10


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 10 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0101


Decimal number 2.222 222 222 222 223 88 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0101

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100